Question

我正在尝试使用Array字段对String进行排序，但排序错误。

我的代码如下。

 let tempWEArray = [
    {
      "from" : "09/2005",
      "to" : "11/2006"
    }, 
    {
      "from" : "09/2006",
      "to" : "11/2007"
    }, 
    {
      "from" : "12/2007",
      "to" : "01/2009"
    }, 
    {
      "from" : "01/2009",
      "to" : "12/2012"
    }, 
    {
      "from" : "01/2013",
      "to" : "03/2018"
    }]

    function sortBy(prop){
        return function(a,b){
            if( a[prop] < b[prop])
            {
                return -1;
            }
            else if( a[prop] > b[prop] )
            {
                return 1;
            }
            return 0;
        }
    }
    
    console.log(tempWEArray.sort(sortBy("to")))

获得的输出如下。

0: Object { from: "12/2007", to: "01/2009" }

1: Object { from: "01/2013", to: "03/2018" }

2: Object { from: "09/2005", to: "11/2006" }

3: Object { from: "09/2006", to: "11/2007" }

4: Object { from: "01/2009", to: "12/2012" }

如上所示，数组未正确排序。一个字段放错了位置。我做错什么了吗？

以下所有答案均有效，我已经选择了已实现的答案。谢谢大家。

Answer 1

您可以先解析这些日期，然后使用<AndroidEnableDesugar>true</AndroidEnableDesugar>对其进行排序。

Answer 2

将其转换为您喜欢的日期，即可按预期工作。

分离日期字符串的每个部分并反转

const dateArray = b[prop].split("/").reverse()

使用传播运算符使用Date.UTC创建时间戳，然后使用new Date创建日期。

new Date(Date.UTC(...dateArray))

然后在两个日期之间使用-来查找排序函数中哪个更大。

一些示例：

const res = new Date(Date.UTC(..."11/2006".split("/").reverse()))

console.log(res);

完整解决方案：

let tempWEArray = [{"from":"09/2005","to":"11/2006"},{"from":"09/2006","to":"11/2007"},{"from":"12/2007","to": "01/2009"},{"from":"01/2009","to": "12/2012"},{"from":"01/2013","to": "03/2018"}]

function sortBy(prop) {
  return function(a, b) {
  
    const dateArr1 = a[prop].split("/").reverse();
    const dateArr2 = b[prop].split("/").reverse();
    
    //make sure months are between 0 and 11
    //can be skipped if this is already ensured.
    dateArr1[1]--;
    dateArr2[1]--;

    return new Date(Date.UTC(...dateArr2)) - new Date(Date.UTC(...dateArr1));
  }
}

console.log(tempWEArray.sort(sortBy("to")))

Answer 3

您可以使用moment.js进行日期露营。例如：

let tempWEArray = [
{
  "from" : "09/2005",
  "to" : "11/2006"
}, 
{
  "from" : "09/2006",
  "to" : "11/2007"
}, 
{
  "from" : "12/2007",
  "to" : "01/2009"
}, 
{
  "from" : "01/2009",
  "to" : "12/2012"
}, 
{
  "from" : "01/2013",
  "to" : "03/2018"
}];

const sortedArray = tempWEArray.sort(
(first, second)  => moment(first.to, 'MM/YYYY')
                    .isSameOrAfter(moment(second.to, 'MM/YYYY')));
console.log(sortedArray);

<script src="https://cdnjs.cloudflare.com/ajax/libs/moment.js/2.23.0/moment.min.js"></script>

Answer 4

function sortBy(prop) {
  return function(a, b) {
const dateArray = b[prop].split("/").reverse()
console.log()
  }
}

Answer 5

您可以用可比较的字符串替换日期模式，如果该字符串没有日期模式，则可以省略。

function sortBy(prop){
    return function(a,b){
        var valueA = a[prop].replace(/^(\d{2})\/(\d{4})$/, '$2-$1'),
            valueB = b[prop].replace(/^(\d{2})\/(\d{4})$/, '$2-$1');

        return valueA.localeCompare(valueB);
    }
}

var tempWEArray = [{ from: "09/2005", to: "11/2006" }, { from: "09/2006", to: "11/2007" }, { from: "12/2007", to: "01/2009" }, { from: "01/2009", to: "12/2012" }, { from: "01/2013", to: "03/2018" }]
   
console.log(tempWEArray.sort(sortBy("to")))

.as-console-wrapper { max-height: 100% !important; top: 0; }

基于字符串值的数组排序不起作用

5 个答案: